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This track of the course covers the topic "Segment Trees".

In details, this track will cover:

  • Basics: Introduction to Segment Trees and when to use them.
  • Approaches:  Understanding how to apply segment-trees in questions.

Objective: The objective of this track is to familiarize the learners with Segment Trees.

Track Content:

  • 1 Video Lecture on Segment Trees.
  • Theoretical Articles.
  • Programming practice problems.

Assessment: All Tracks in every week are associated with weekly contests.

We have combined Classroom and Theory tab and created a new Learn tab for easy access. You can access Classroom and Theory from the left panel.

In this video, segment tree is introduced with an example problem.
Codes:


In this video segment tree construction is discussed with detailed algorithm.
Codes:


In this video, getSum() query for given array is discussed.
 Codes:


In this video update query is discussed with implementation.
Codes:


Binary Indexed Tree (Intoductiion)

Binary Indexed Tree (An Example Problem)

Binary Indxed Tree (Prefix Sum)
Segment Trees are Binary Tree which is used to store intervals or segments. That is each node in a segment tree basically stores the segment of an array. Segment Trees are generally used in problems where we need to solve queries on a range of elements in arrays.

Let us consider the following problem to understand Segment Trees.

Problem: We have an array arr[0 . . . n-1]. We should be able to perform the below operations on the array:
  1. Find the sum of elements from index l to r where 0 <= l <= r <= n-1.
  2. Change value of a specified element of the array to a new value x. We need to do arr[i] = x where 0 <= i <= n-1.

General Solution: A simple solution is to run a loop from index l to r and calculate the sum of elements in the given range. To update a value, simply do arr[i] = x. The first operation takes O(N) time for every query, where N is the number of elements in the range [l,r] and the second operation takes O(1) time.

Can we optimize the time complexity of the first operation in the above solution?


Yes, we can optimize the first operation to be solved in O(1) time complexity by storing presum. We can keep an auxiliary array say sum[] in which the i-th element will store the sum of first i elements of the original array. So, whenever we need to find the sum of a range of elements, we can simply calculate it by (sum[r]-sum[l-1]). But in this solution the complexity to perform the second operation of updating an element increases from O(1) to O(N).

What if the number of query and updates are equal? Can we perform both the operations in O(log n) time once given the array?


We can use a segment tree to perform both of the operations in O(log N) time complexity.

Representation of Segment trees:
  1. Leaf Nodes are the elements of the input array.
  2. Each internal node represents some merging of the leaf nodes. The merging may be different for different problems. For this problem, merging is the sum of leaves under a node.

An array representation of the tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i + 1, right child at 2*i + 2 and the parent is at (i - 1)/2.



Array representation of Segment Trees: Like Heap, segment tree is also represented as an array. The difference here is, it is not a complete binary tree. It is rather a full binary tree (every node has 0 or 2 children) and all levels are filled except possibly the last level. Unlike Heap, the last level may have gaps between nodes. Below are the values in the segment tree array for the above diagram.

Array representation of segment tree for input array {1, 3, 5, 7, 9, 11} is,

st[] = {36, 9, 27, 4, 5, 16, 11, 1, 3, DUMMY, DUMMY, 7, 9, DUMMY, DUMMY}

The dummy values are never accessed and have no use. This is some wastage of space due to simple array representation. We may optimize this wastage using some clever implementations, but the code for sum and update becomes more complex.

Construction of Segment Tree from the given array: We start with a segment arr[0 . . . n-1] and every time we divide the current segment into two halves(if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment, we store the sum in the corresponding node.

All levels of the constructed segment tree will be completely filled except the last level. Also, the tree will be a Full Binary Tree because we always divide segments into two halves at every level. Since the constructed tree is always a full binary tree with n leaves, there will be n-1 internal nodes. So the total number of nodes will be 2*n – 1. Note that this does not include dummy nodes.

What is the total size of the array representing segment tree? If n is a power of 2, then there are no dummy nodes. So the size of the segment tree is 2n-1 (n leaf nodes and n-1) internal nodes. If n is not a power of 2, the size of the tree will be 2*x – 1 where x is the smallest of 2 greater than n. For example, when n = 10, then size of array representing segment tree is 2*16-1 = 31.

Query for Sum of given range

Following is the algorithm to get the sum of elements.
int getSum(node, l, r) 
{
if the range of the node is within l and r
return value in the node
else if the range of the node is completely outside l and r
return 0
else
return getSum(node's left child, l, r) +
getSum(node's right child, l, r)
}

Updating a value

Like tree construction and query operations, the update can also be done recursively. We are given an index which needs to be updated. Let diff be the value to be added. We start from the root of the segment tree and add diff to all nodes which have given index in their range. If a node doesn’t have given index in its range, we don’t make any changes to that node.

Implementation:
// C++ program to show segment tree operations like construction, query
// and update
#include <bits/stdc++.h>
using namespace std;
// A utility function to get the middle index from corner indexes.
int getMid(int s, int e) { return s + (e -s)/2; }
/* A recursive function to get the sum of values in given range
of the array. The following are parameters for this function.
st --> Pointer to segment tree
si --> Index of current node in the segment tree. Initially
0 is passed as root is always at index 0
ss & se --> Starting and ending indexes of the segment represented
by current node, i.e., st[si]
qs & qe --> Starting and ending indexes of query range */
int getSumUtil(int *st, int ss, int se, int qs, int qe, int si)
{
// If segment of this node is a part of given range, then return
// the sum of the segment
if (qs <= ss && qe >= se)
return st[si];
// If segment of this node is outside the given range
if (se < qs || ss > qe)
return 0;
// If a part of this segment overlaps with the given range
int mid = getMid(ss, se);
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// Java Program to show segment tree operations like construction,
// query and update
class SegmentTree
{
int st[]; // The array that stores segment tree nodes
/* Constructor to construct segment tree from given array. This
constructor allocates memory for segment tree and calls
constructSTUtil() to fill the allocated memory */
SegmentTree(int arr[], int n)
{
// Allocate memory for segment tree
//Height of segment tree
int x = (int) (Math.ceil(Math.log(n) / Math.log(2)));
//Maximum size of segment tree
int max_size = 2 * (int) Math.pow(2, x) - 1;
st = new int[max_size]; // Memory allocation
constructSTUtil(arr, 0, n - 1, 0);
}
// A utility function to get the middle index from corner indexes.
int getMid(int s, int e) {
return s + (e - s) / 2;
}
/* A recursive function to get the sum of values in given range
of the array. The following are parameters for this function.
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Complete Code/Output:
// C++ program to show segment tree operations like construction, query  
// and update  
#include <bits/stdc++.h> 
using namespace std; 
  
// A utility function to get the middle index from corner indexes.  
int getMid(int s, int e) { return s + (e -s)/2; }  
  
/* A recursive function to get the sum of values in given range  
    of the array. The following are parameters for this function.  
  
    st --> Pointer to segment tree  
    si --> Index of current node in the segment tree. Initially  
            0 is passed as root is always at index 0  
    ss & se --> Starting and ending indexes of the segment represented  
                by current node, i.e., st[si]  
    qs & qe --> Starting and ending indexes of query range */
int getSumUtil(int *st, int ss, int se, int qs, int qe, int si)  
{  
    // If segment of this node is a part of given range, then return  
    // the sum of the segment  
    if (qs <= ss && qe >= se)  
        return st[si];  
  
    // If segment of this node is outside the given range  
    if (se < qs || ss > qe)  
        return 0;  
  
    // If a part of this segment overlaps with the given range  
    int mid = getMid(ss, se);  
    return getSumUtil(st, ss, mid, qs, qe, 2*si+1) +  
        getSumUtil(st, mid+1, se, qs, qe, 2*si+2);  
}  
  
/* A recursive function to update the nodes which have the given  
index in their range. The following are parameters  
    st, si, ss and se are same as getSumUtil()  
    i --> index of the element to be updated. This index is  
            in the input array.  
diff --> Value to be added to all nodes which have i in range */
void updateValueUtil(int *st, int ss, int se, int i, int diff, int si)  
{  
    // Base Case: If the input index lies outside the range of  
    // this segment  
    if (i < ss || i > se)  
        return;  
  
    // If the input index is in range of this node, then update  
    // the value of the node and its children  
    st[si] = st[si] + diff;  
    if (se != ss)  
    {  
        int mid = getMid(ss, se);  
        updateValueUtil(st, ss, mid, i, diff, 2*si + 1);  
        updateValueUtil(st, mid+1, se, i, diff, 2*si + 2);  
    }  
}  
  
// The function to update a value in input array and segment tree.  
// It uses updateValueUtil() to update the value in segment tree  
void updateValue(int arr[], int *st, int n, int i, int new_val)  
{  
    // Check for erroneous input index  
    if (i < 0 || i > n-1)  
    {  
        cout<<"Invalid Input";  
        return;  
    }  
  
    // Get the difference between new value and old value  
    int diff = new_val - arr[i];  
  
    // Update the value in array  
    arr[i] = new_val;  
  
    // Update the values of nodes in segment tree  
    updateValueUtil(st, 0, n-1, i, diff, 0);  
}  
  
// Return sum of elements in range from index qs (quey start)  
// to qe (query end). It mainly uses getSumUtil()  
int getSum(int *st, int n, int qs, int qe)  
{  
    // Check for erroneous input values  
    if (qs < 0 || qe > n-1 || qs > qe)  
    {  
        cout<<"Invalid Input";  
        return -1;  
    }  
  
    return getSumUtil(st, 0, n-1, qs, qe, 0);  
}  
  
// A recursive function that constructs Segment Tree for array[ss..se].  
// si is index of current node in segment tree st  
int constructSTUtil(int arr[], int ss, int se, int *st, int si)  
{  
    // If there is one element in array, store it in current node of  
    // segment tree and return  
    if (ss == se)  
    {  
        st[si] = arr[ss];  
        return arr[ss];  
    }  
  
    // If there are more than one elements, then recur for left and  
    // right subtrees and store the sum of values in this node  
    int mid = getMid(ss, se);  
    st[si] = constructSTUtil(arr, ss, mid, st, si*2+1) +  
            constructSTUtil(arr, mid+1, se, st, si*2+2);  
    return st[si];  
}  
  
/* Function to construct segment tree from given array. This function  
allocates memory for segment tree and calls constructSTUtil() to  
fill the allocated memory */
int *constructST(int arr[], int n)  
{  
    // Allocate memory for the segment tree  
  
    //Height of segment tree  
    int x = (int)(ceil(log2(n)));  
  
    //Maximum size of segment tree  
    int max_size = 2*(int)pow(2, x) - 1;  
  
    // Allocate memory  
    int *st = new int[max_size];  
  
    // Fill the allocated memory st  
    constructSTUtil(arr, 0, n-1, st, 0);  
  
    // Return the constructed segment tree  
    return st;  
}  
  
// Driver program to test above functions  
int main()  
{  
    int arr[] = {1, 3, 5, 7, 9, 11};  
    int n = sizeof(arr)/sizeof(arr[0]);  
  
    // Build segment tree from given array  
    int *st = constructST(arr, n);  
  
    // Print sum of values in array from index 1 to 3  
    cout<<"Sum of values in given range = "<<getSum(st, n, 1, 3)<<endl;  
  
    // Update: set arr[1] = 10 and update corresponding  
    // segment tree nodes  
    updateValue(arr, st, n, 1, 10);  
  
    // Find sum after the value is updated  
    cout<<"Updated sum of values in given range = "
            <<getSum(st, n, 1, 3)<<endl;  
    return 0;  
} 

// Java Program to show segment tree operations like construction, 
// query and update 
class SegmentTree  
{ 
    int st[]; // The array that stores segment tree nodes 
  
    /* Constructor to construct segment tree from given array. This 
       constructor  allocates memory for segment tree and calls 
       constructSTUtil() to  fill the allocated memory */
    SegmentTree(int arr[], int n) 
    { 
        // Allocate memory for segment tree 
        //Height of segment tree 
        int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); 
  
        //Maximum size of segment tree 
        int max_size = 2 * (int) Math.pow(2, x) - 1; 
  
        st = new int[max_size]; // Memory allocation 
  
        constructSTUtil(arr, 0, n - 1, 0); 
    } 
  
    // A utility function to get the middle index from corner indexes. 
    int getMid(int s, int e) { 
        return s + (e - s) / 2; 
    } 
  
    /*  A recursive function to get the sum of values in given range 
        of the array.  The following are parameters for this function. 
  
      st    --> Pointer to segment tree 
      si    --> Index of current node in the segment tree. Initially 
                0 is passed as root is always at index 0 
      ss & se  --> Starting and ending indexes of the segment represented 
                    by current node, i.e., st[si] 
      qs & qe  --> Starting and ending indexes of query range */
    int getSumUtil(int ss, int se, int qs, int qe, int si) 
    { 
        // If segment of this node is a part of given range, then return 
        // the sum of the segment 
        if (qs <= ss && qe >= se) 
            return st[si]; 
  
        // If segment of this node is outside the given range 
        if (se < qs || ss > qe) 
            return 0; 
  
        // If a part of this segment overlaps with the given range 
        int mid = getMid(ss, se); 
        return getSumUtil(ss, mid, qs, qe, 2 * si + 1) + 
                getSumUtil(mid + 1, se, qs, qe, 2 * si + 2); 
    } 
  
    /* A recursive function to update the nodes which have the given  
       index in their range. The following are parameters 
        st, si, ss and se are same as getSumUtil() 
        i    --> index of the element to be updated. This index is in 
                 input array. 
       diff --> Value to be added to all nodes which have i in range */
    void updateValueUtil(int ss, int se, int i, int diff, int si) 
    { 
        // Base Case: If the input index lies outside the range of  
        // this segment 
        if (i < ss || i > se) 
            return; 
  
        // If the input index is in range of this node, then update the 
        // value of the node and its children 
        st[si] = st[si] + diff; 
        if (se != ss) { 
            int mid = getMid(ss, se); 
            updateValueUtil(ss, mid, i, diff, 2 * si + 1); 
            updateValueUtil(mid + 1, se, i, diff, 2 * si + 2); 
        } 
    } 
  
    // The function to update a value in input array and segment tree. 
   // It uses updateValueUtil() to update the value in segment tree 
    void updateValue(int arr[], int n, int i, int new_val) 
    { 
        // Check for erroneous input index 
        if (i < 0 || i > n - 1) { 
            System.out.println("Invalid Input"); 
            return; 
        } 
  
        // Get the difference between new value and old value 
        int diff = new_val - arr[i]; 
  
        // Update the value in array 
        arr[i] = new_val; 
  
        // Update the values of nodes in segment tree 
        updateValueUtil(0, n - 1, i, diff, 0); 
    } 
  
    // Return sum of elements in range from index qs (quey start) to 
   // qe (query end).  It mainly uses getSumUtil() 
    int getSum(int n, int qs, int qe) 
    { 
        // Check for erroneous input values 
        if (qs < 0 || qe > n - 1 || qs > qe) { 
            System.out.println("Invalid Input"); 
            return -1; 
        } 
        return getSumUtil(0, n - 1, qs, qe, 0); 
    } 
  
    // A recursive function that constructs Segment Tree for array[ss..se]. 
    // si is index of current node in segment tree st 
    int constructSTUtil(int arr[], int ss, int se, int si) 
    { 
        // If there is one element in array, store it in current node of 
        // segment tree and return 
        if (ss == se) { 
            st[si] = arr[ss]; 
            return arr[ss]; 
        } 
  
        // If there are more than one elements, then recur for left and 
        // right subtrees and store the sum of values in this node 
        int mid = getMid(ss, se); 
        st[si] = constructSTUtil(arr, ss, mid, si * 2 + 1) + 
                 constructSTUtil(arr, mid + 1, se, si * 2 + 2); 
        return st[si]; 
    } 
  
    // Driver program to test above functions 
    public static void main(String args[]) 
    { 
        int arr[] = {1, 3, 5, 7, 9, 11}; 
        int n = arr.length; 
        SegmentTree  tree = new SegmentTree(arr, n); 
  
        // Build segment tree from given array 
  
        // Print sum of values in array from index 1 to 3 
        System.out.println("Sum of values in given range = " + 
                           tree.getSum(n, 1, 3)); 
  
        // Update: set arr[1] = 10 and update corresponding segment 
        // tree nodes 
        tree.updateValue(arr, n, 1, 10); 
  
        // Find sum after the value is updated 
        System.out.println("Updated sum of values in given range = " + 
                tree.getSum(n, 1, 3)); 
    } 
} 

Sum of values in given range = 15
Updated sum of values in given range = 22

Time Complexity: The time Complexity for tree construction is O(n). There are total 2n-1 nodes, and the value of every node is calculated only once in tree construction.

Time complexity to query is O(Logn). To query a sum, we process at most four nodes at every level and number of levels is O(Logn).

The time complexity of the update is also O(Logn). To update a leaf value, we process one node at every level and number of levels is O(Logn).
The Range Minimum Query is another popular problem which can be solved using Segment Trees. The problems state that, given an array and a list of queries containing ranges, the task is to find the minimum element in the range for every query.

Note: This problem does not require us to perform any update operation on the array.

Representation of Segment trees:
  1. Leaf Nodes are the elements of the input array.
  2. Each internal node represents a minimum of all leaves under it.

An array representation of the tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i + 1, the right child at 2*i + 2 and the parent is at (i-1)/2.



We start with a segment arr[0 . . . n-1] and every time we divide the current segment into two halves(if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment, we store the minimum value in a segment tree node.

All levels of the constructed segment tree will be completely filled except the last level. Also, the tree will be a Full Binary Tree because we always divide segments into two halves at every level. Since the constructed tree is always a full binary tree with n leaves, there will be n-1 internal nodes. So the total number of nodes will be 2*n – 1.

Height of the segment tree will be log2N. Since the tree is represented using array and relation between parent and child indexes must be maintained, the size of memory allocated for segment tree will be 2*2log2N - 1.

Query for the minimum value in a given range:
Following is the algorithm to get the minimum element in Range.
// qs --> query start index, qe --> query end index
int RMQ(node, qs, qe)
{
if the range of node is within qs and qe
return value in the node
else if the range of node is completely outside qs and qe
return INFINITE
else
return min( RMQ(node's left child, qs, qe),
RMQ(node's right child, qs, qe) )
}

Implementation:

// C++ program for range minimum
// query using segment tree
#include <bits/stdc++.h>
using namespace std;
// A utility function to get minimum of two numbers
int minVal(int x, int y) { return (x < y)? x: y; }
// A utility function to get the
// middle index from corner indexes.
int getMid(int s, int e) { return s + (e -s)/2; }
/* A recursive function to get the
minimum value in a given range
of array indexes. The following
are parameters for this function.
st --> Pointer to segment tree
index --> Index of current node in the
segment tree. Initially 0 is
passed as root is always at index 0
ss & se --> Starting and ending indexes
of the segment represented
by current node, i.e., st[index]
qs & qe --> Starting and ending indexes of query range */
int RMQUtil(int *st, int ss, int se, int qs, int qe, int index)
{
// If segment of this node is a part
// of given range, then return
// the min of the segment
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// Program for range minimum query using segment tree
class SegmentTreeRMQ
{
int st[]; //array to store segment tree
// A utility function to get minimum of two numbers
int minVal(int x, int y) {
return (x < y) ? x : y;
}
// A utility function to get the middle index from corner
// indexes.
int getMid(int s, int e) {
return s + (e - s) / 2;
}
/* A recursive function to get the minimum value in a given
range of array indexes. The following are parameters for
this function.
st --> Pointer to segment tree
index --> Index of current node in the segment tree. Initially
0 is passed as root is always at index 0
ss & se --> Starting and ending indexes of the segment
represented by current node, i.e., st[index]
qs & qe --> Starting and ending indexes of query range */
int RMQUtil(int ss, int se, int qs, int qe, int index)
{
// If segment of this node is a part of given range, then
// return the min of the segment
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Complete Code/Output:
// C++ program for range minimum 
// query using segment tree  
#include <bits/stdc++.h> 
using namespace std; 
  
// A utility function to get minimum of two numbers  
int minVal(int x, int y) { return (x < y)? x: y; }  
  
// A utility function to get the  
// middle index from corner indexes.  
int getMid(int s, int e) { return s + (e -s)/2; }  
  
/* A recursive function to get the 
minimum value in a given range  
of array indexes. The following  
are parameters for this function.  
  
    st --> Pointer to segment tree  
    index --> Index of current node in the  
           segment tree. Initially 0 is  
           passed as root is always at index 0  
    ss & se --> Starting and ending indexes  
                of the segment represented  
                by current node, i.e., st[index]  
    qs & qe --> Starting and ending indexes of query range */
int RMQUtil(int *st, int ss, int se, int qs, int qe, int index)  
{  
    // If segment of this node is a part  
    // of given range, then return  
    // the min of the segment  
    if (qs <= ss && qe >= se)  
        return st[index];  
  
    // If segment of this node 
    // is outside the given range  
    if (se < qs || ss > qe)  
        return INT_MAX;  
  
    // If a part of this segment 
    // overlaps with the given range  
    int mid = getMid(ss, se);  
    return minVal(RMQUtil(st, ss, mid, qs, qe, 2*index+1),  
                RMQUtil(st, mid+1, se, qs, qe, 2*index+2));  
}  
  
// Return minimum of elements in range 
// from index qs (quey start) to  
// qe (query end). It mainly uses RMQUtil()  
int RMQ(int *st, int n, int qs, int qe)  
{  
    // Check for erroneous input values  
    if (qs < 0 || qe > n-1 || qs > qe)  
    {  
        cout<<"Invalid Input";  
        return -1;  
    }  
  
    return RMQUtil(st, 0, n-1, qs, qe, 0);  
}  
  
// A recursive function that constructs 
// Segment Tree for array[ss..se].  
// si is index of current node in segment tree st  
int constructSTUtil(int arr[], int ss, int se, 
                                int *st, int si)  
{  
    // If there is one element in array, 
    // store it in current node of  
    // segment tree and return  
    if (ss == se)  
    {  
        st[si] = arr[ss];  
        return arr[ss];  
    }  
  
    // If there are more than one elements,  
    // then recur for left and right subtrees  
    // and store the minimum of two values in this node  
    int mid = getMid(ss, se);  
    st[si] = minVal(constructSTUtil(arr, ss, mid, st, si*2+1),  
                    constructSTUtil(arr, mid+1, se, st, si*2+2));  
    return st[si];  
}  
  
/* Function to construct segment tree  
from given array. This function allocates 
memory for segment tree and calls constructSTUtil() to  
fill the allocated memory */
int *constructST(int arr[], int n)  
{  
    // Allocate memory for segment tree  
  
    //Height of segment tree  
    int x = (int)(ceil(log2(n)));  
  
    // Maximum size of segment tree  
    int max_size = 2*(int)pow(2, x) - 1;  
  
    int *st = new int[max_size];  
  
    // Fill the allocated memory st  
    constructSTUtil(arr, 0, n-1, st, 0);  
  
    // Return the constructed segment tree  
    return st;  
}  
  
// Driver Code  
int main()  
{  
    int arr[] = {1, 3, 2, 7, 9, 11};  
    int n = sizeof(arr)/sizeof(arr[0]);  
  
    // Build segment tree from given array  
    int *st = constructST(arr, n);  
  
    int qs = 1; // Starting index of query range  
    int qe = 5; // Ending index of query range  
  
    // Print minimum value in arr[qs..qe]  
    cout<<"Minimum of values in range ["<<qs<<", "<<qe<<"] "<< 
    "is = "<<RMQ(st, n, qs, qe)<<endl;  
  
    return 0;  
} 

// Program for range minimum query using segment tree 
class SegmentTreeRMQ 
{ 
    int st[]; //array to store segment tree 
  
    // A utility function to get minimum of two numbers 
    int minVal(int x, int y) { 
        return (x < y) ? x : y; 
    } 
  
    // A utility function to get the middle index from corner 
    // indexes. 
    int getMid(int s, int e) { 
        return s + (e - s) / 2; 
    } 
  
    /*  A recursive function to get the minimum value in a given 
        range of array indexes. The following are parameters for 
        this function. 
  
        st    --> Pointer to segment tree 
        index --> Index of current node in the segment tree. Initially 
                   0 is passed as root is always at index 0 
        ss & se  --> Starting and ending indexes of the segment 
                     represented by current node, i.e., st[index] 
        qs & qe  --> Starting and ending indexes of query range */
    int RMQUtil(int ss, int se, int qs, int qe, int index) 
    { 
        // If segment of this node is a part of given range, then 
        // return the min of the segment 
        if (qs <= ss && qe >= se) 
            return st[index]; 
  
        // If segment of this node is outside the given range 
        if (se < qs || ss > qe) 
            return Integer.MAX_VALUE; 
  
        // If a part of this segment overlaps with the given range 
        int mid = getMid(ss, se); 
        return minVal(RMQUtil(ss, mid, qs, qe, 2 * index + 1), 
                RMQUtil(mid + 1, se, qs, qe, 2 * index + 2)); 
    } 
  
    // Return minimum of elements in range from index qs (quey 
    // start) to qe (query end).  It mainly uses RMQUtil() 
    int RMQ(int n, int qs, int qe) 
    { 
        // Check for erroneous input values 
        if (qs < 0 || qe > n - 1 || qs > qe) { 
            System.out.println("Invalid Input"); 
            return -1; 
        } 
  
        return RMQUtil(0, n - 1, qs, qe, 0); 
    } 
  
    // A recursive function that constructs Segment Tree for 
    // array[ss..se]. si is index of current node in segment tree st 
    int constructSTUtil(int arr[], int ss, int se, int si) 
    { 
        // If there is one element in array, store it in current 
        //  node of segment tree and return 
        if (ss == se) { 
            st[si] = arr[ss]; 
            return arr[ss]; 
        } 
  
        // If there are more than one elements, then recur for left and 
        // right subtrees and store the minimum of two values in this node 
        int mid = getMid(ss, se); 
        st[si] = minVal(constructSTUtil(arr, ss, mid, si * 2 + 1), 
                constructSTUtil(arr, mid + 1, se, si * 2 + 2)); 
        return st[si]; 
    } 
  
    /* Function to construct segment tree from given array. This function 
       allocates memory for segment tree and calls constructSTUtil() to 
       fill the allocated memory */
    void constructST(int arr[], int n) 
    { 
        // Allocate memory for segment tree 
  
        //Height of segment tree 
        int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); 
  
        // Maximum size of segment tree 
        int max_size = 2 * (int) Math.pow(2, x) - 1; 
        st = new int[max_size]; // allocate memory 
  
        // Fill the allocated memory st 
        constructSTUtil(arr, 0, n - 1, 0); 
    } 
  
    // Driver Code 
    public static void main(String args[])  
    { 
        int arr[] = {1, 3, 2, 7, 9, 11}; 
        int n = arr.length; 
        SegmentTreeRMQ tree = new SegmentTreeRMQ(); 
  
        // Build segment tree from given array 
        tree.constructST(arr, n); 
  
        int qs = 1;  // Starting index of query range 
        int qe = 5;  // Ending index of query range 
  
        // Print minimum value in arr[qs..qe] 
        System.out.println("Minimum of values in range [" + qs + ", "
                           + qe + "] is = " + tree.RMQ(n, qs, qe)); 
    } 
} 

Minimum of values in range [1, 5] is = 2

Time Complexity: The time complexity for tree construction is O(n). There are total 2n-1 nodes, and the value of every node is calculated only once in tree construction.

Time complexity to query is O(Logn). To query a range minimum, we process at most two nodes at every level and number of levels is O(Logn).

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